Change in Angular Velocity?

Question

A merry-go-round of radius R, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round:
1.) m =10kg, r=.50
2.) m =40kg, r=.25
3.) m =10kg, r=.1.0
4.) m =20kg, r=.25
5.) m =10kg, r=.25
6.) m =15kg, r=.75

Answer 1

2
5
4
1
3
6

Answer 2

P = mv^2/r = m w^2 r; where P is the centripetal/centrifugal force that is exactly offset by the frictional force F = k mg since the bag "sticks" each time. k is the friction coefficient, m the bag mass, and g is 9.81 m/sec^2 on Earth's surface. Thus, F = k mg = m w^2 r = P, and w = the angular velocity of the MGR you are looking for.

We can write w^2 = k g/r; so if we do a ratio of two angular velocities (W/w)^2 = (k mg/R)/(k mg/r) = r/R, we see that relative angular velocity depends solely on the radius where the bag was dropped. That is W^2 = w^2 (r/R). Mass has no bearing on it so long as the bag "sticks" each time.

Thus, rank the radii (1,.75,.5,.25,.25,.25) and that's the increasing rank order (highest to the right) of the angular velocities. For example, if r = 1 and R = .75, then W^2 = w^2 (1/.75) so that W(.75) > w(1), which is to say the angular velocity of the closer in radius (R = .75) will be greater than that of the farther out (r = 1.0).

If you've ever done the conservation of angular momentum experiment by rotating on a swivel chair and pulling in your arms to see the angular velocity increase, you'll understand the results here...same thing. Angular velocity increases with decreasing r.

Answer 3

From conservation of angular momentum,
I1ω1 = (I1 + I2)ω2
This means that the larger the added moment of inertia I2 about the center of rotation, the smaller will be the angular rate ω2 after the addition of the sandbag. The added moment of inertia
I2 = mr^2
So just work out the 6 values for I; if you rank them in decreasing order you are ranking final angular rate in increasing order.
I think you'll find disagreement with the 1st answer. The 2nd answer is mistaken; it disregards the sandbag mass that must be accelerated from 0 to the final speed, and gets involved with friction which in the problem statement is clearly irrelevant.